Free Pearson Edexcel International A Level Physics lesson

Investigate motion and materials

Plan, analyse and improve core practicals 1–3.

What you’ll learn

  • Analyse a free-fall measurement without concealing release offsets.
  • Verify terminal speed before determining viscosity.
  • Measure small elastic extensions and distinguish stiffness from Young modulus.

Core practical 1: determine gravitational acceleration

For an object released from rest with negligible drag, vertical displacement s obeys s = ½gt². A timed drop can therefore determine g from distance and time, or more reliably from the gradient of a graph of s against t² across several distances. Choose an electronic release and timing arrangement whose start and stop events correspond clearly to the measured displacement. Keep the drop path clear and use a suitable catch area.

A light-gate alternative measures speeds at known positions using an interrupting card of measured length, then relates v² to displacement. The card length divided by blocking time estimates an average speed over the gate passage; it approximates instantaneous speed only when the speed changes little across that interval. State the chosen method rather than mixing a release-time model with an unrelated gate-time reading.

Measure the actual travel distance between the timing events and repeat drops at several heights. A release delay, residual initial velocity or an incorrect distance origin can bias the result. Electronic timing reduces human reaction effects but does not automatically remove trigger offsets. Inspect the graph’s intercept and residuals, and avoid a very light object for which air resistance is significant. Report the fitted g with a discussion of the dominant uncertainty and the validity of the rest-release assumption.

Rest release: s = ½gt²; slope of s against t² = g/2. Alternative: v² = u² + 2gs.

Symbols and units

  • s is downward displacement in m
  • u and v are downward velocities in m s⁻¹
  • t is time in s
  • g is acceleration magnitude in m s⁻².

How the equation explains the idea

  1. Release from rest with negligible drag gives s = gt²/2. On s-versus-t² axes the gradient equals g/2, not g.
  2. If the initial speed is not zero, v² = u² + 2gs provides an alternative linear plot of v² against s: gradient 2g and intercept u². Timing offsets and drag are still possible systematic limitations.
The transformed fall graph has gradient g/2

Conditions. Ideal release from rest with downward acceleration g = 9.81 m s⁻², no drag and exact time zero. The horizontal variable is t², not t.

Time squared / s²; Downward displacement / m.

The model s = 4.905t² gives 0.981 m at t² = 0.20 s². Gradient is 4.905 m s⁻², so twice the gradient gives g. A curved displacement–time graph becomes a line on these transformed axes.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Interpret a drop graph

  1. A fitted s-versus-t² graph has slope 4.92 m s⁻².
  2. The model identifies this slope as g/2.
  3. Therefore g = 9.84 m s⁻²; a nonzero intercept still requires checking separately.

The slope is half g, not g itself.

Keep this idea. Align the measured distance with the timing events and test the release model using a graph.

Core practical 2: a falling sphere in liquid

A small sphere falling through a liquid experiences weight downward and upthrust plus viscous drag upward. At terminal speed these forces balance. For slow laminar flow around a small sphere, Stokes’ law gives drag 6πηrv. Combining the force balance with the sphere volume gives η = 2r²g(ρsphere − ρliquid)/(9v). The result depends on radius squared and on the measured speed actually being terminal.

Use a sufficiently tall and wide transparent column, mark a timing region below an initial settling distance and measure times across successive equal segments. If speeds agree within uncertainty, choose a suitable terminal region and repeat drops. Measure sphere diameter in several orientations, determine or use justified densities, and keep liquid temperature controlled because viscosity changes with temperature. Avoid large spheres or narrow columns that produce significant wall effects or turbulent motion.

The practical method needs care in releasing the sphere near the column centre without an extra push and in identifying when its centre crosses each mark. Video or a suitable sensor can reduce manual timing uncertainty but still requires scale and time calibration. Use spill containment and a stable column, and retrieve spheres safely without creating a slipping or glass-breakage hazard. Evaluate which measured quantity or assumption most limits the result before recommending simply more repeats.

At terminal speed: weight = upthrust + drag; η = 2r²g(ρsphere − ρliquid)/(9v), within Stokes’ regime.

Symbols and units

  • r is sphere radius in m
  • ρsphere and ρliquid are densities in kg m⁻³
  • v is terminal speed in m s⁻¹
  • η is dynamic viscosity in Pa s
  • g is acceleration in m s⁻².

How the equation explains the idea

  1. At terminal speed, excess weight after buoyancy is (4πr³/3)(ρsphere − ρliquid)g. Set this equal to Stokes drag 6πηrv.
  2. Cancel common factors and solve for η = 2r²g(ρsphere − ρliquid)/(9v). This needs low-Reynolds-number flow, terminal motion and negligible wall effects; radius uncertainty enters twice in relative terms.

Worked example

Check terminal motion before viscosity

  1. A sphere crosses successive 0.10 m regions in 2.0 s, 2.0 s and 2.0 s within timing uncertainty.
  2. Each speed is 0.10/2.0 = 0.050 m s⁻¹, supporting a terminal region.
  3. Use this speed with measured radius and densities only if the small-sphere laminar and negligible-wall-effect assumptions are also credible.

Agreement of segment speeds is necessary evidence for the force balance, though not its only assumption.

Keep this idea. Verify terminal motion, control temperature and justify the Stokes-law conditions before calculating viscosity.

Core practical 3: a material property from a small extension

Young modulus is the ratio of tensile stress to tensile strain in the linear elastic region. Stress is force divided by original cross-sectional area, and strain is extension divided by original gauge length. The resulting relation E = FL/(AΔL) shows why a force–extension experiment alone measures specimen stiffness rather than the material modulus. The original length and area are essential additional measurements.

Secure a long thin wire in a stable arrangement and apply safe incremental loads. Use a small preload to remove slack, recording the corresponding reference length or reading consistently. Measure small extensions with a suitable displacement gauge, micrometer arrangement or other calibrated method, and measure diameter in several positions and orientations. A reference wire or differential arrangement can reduce the influence of common support movement or temperature changes, but its assumptions must be understood.

Take a force–extension series within the safe linear region, check recovery on unloading and obtain stiffness from the fitted gradient. Then E = kL/A. Use a long gauge length to make extension more measurable without simply approaching fracture. Secure suspended masses over a low catch tray and keep clear of the tensioned wire. A residual extension on unloading suggests that the elastic limit was exceeded, in which case those data should not be used indiscriminately to calculate one linear Young modulus.

E = stress/strain = FL/(AΔL); k = ΔF/ΔL in a linear region; E = kL/A.

Symbols and units

  • E is Young modulus in Pa
  • F is tensile force in N
  • L is original gauge length in m
  • A is area in m²
  • ΔL is extension in m
  • k is force–extension gradient in N m⁻¹.

How the equation explains the idea

  1. Divide stress F/A by strain ΔL/L to obtain E = FL/(AΔL). Within the linear region F/ΔL equals fitted gradient k.
  2. Therefore E = kL/A. With k = 20000 N m⁻¹, L = 2.0 m and A = 2.0 × 10⁻⁷ m², E = 2.0 × 10¹¹ Pa. Length and area affect specimen stiffness but not the unchanged material modulus.

Worked example

Convert specimen stiffness to modulus

  1. A fitted stiffness is 2.0 × 10⁴ N m⁻¹ for gauge length 1.50 m.
  2. The original cross-sectional area is 3.0 × 10⁻⁷ m².
  3. Young modulus = kL/A = 1.0 × 10¹¹ Pa.

The modulus follows from stiffness and original geometry together.

Keep this idea. Measure original geometry and small reversible extensions, then convert the linear stiffness into Young modulus.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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